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I seem to have thought myself into a corner. Can someone point out the hole in my reasoning here.

Suppose $f:X \rightarrow \mathbb R$ where $f(x) = \|x\|,$ for $x$ in $X$. Knowing that $\|x\|=0$ implies $x = 0$ we should have $f$ injective since $\operatorname{null}(f) = \{0\}$.

But, by counter-example, we could take $x_1 = (1, 0)$ and $x_2 = (0, 1)$ then we would have $f(x_1) = f(x_2)$ with $x_1 \neq x_2$ contradicting injectivity.

Clearly there is a steaming pile of nonsense hidden here somewhere but it seems to be eluding me.

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up vote 6 down vote accepted

The kernel of a function being $\{0\}$ only implies that the function is injective if it's linear, or an algebraic homomorphism, or something like that. The function we're talking about is not.

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Of course! I knew I was missing something obvious. Thanks for curing my frustration. – Zachary Mar 7 '14 at 4:04

It does not seem to be injective! Trivially, multiple vectors can have the same norm!

For instance, consider any vector $(a_1,a_2).$ Then $(a_2,a_1)$ has the same norm.

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